$\dfrac{d}{dx}[x^3-\cos(x)]=$
Solution: The expression to differentiate includes $\cos(x)$. Remember that the derivative of $\cos(x)$ is $-\sin(x)$. Put another way, $\dfrac{d}{dx}[\cos(x)]=-\sin(x)$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[x^3-\cos(x)] \\\\ &=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}[\cos(x)] \\\\ &=3x^2-(-\sin(x)) \\\\ &=3x^2+\sin(x) \end{aligned}$ In conclusion, $\dfrac{d}{dx}[x^3-\cos(x)]=3x^2+\sin(x)$